随机过程习题答案

1 X ( )与 X (1)的联合分布律为 2 1 X( ) 0 1 2 X (1) −1 +2 1 2 0 0 1 2
0, 0, 1 1 , ⇒ F ( x1 , x2 ; ,1) = 2 2 1 , 2 1,
x1 < 0, −∞ < x2 < +∞ x1 ≥ 0, x2 < −1 0 ≤ x1 < 1, x2 ≥ −1 x1 ≥ 1, −1 ≤ x2 < 2 x1 ≥ 1, x2 ≥ 2
相关函数表示Y (t ) = X (t + a ) − X (t ), t ∈ T的自相关函数.
解
RY (t1 , t 2 ) = E[Y (t1 )Y (t 2 )] = E[( X (t1 + a ) − X (t1 ))( X (t 2 + a ) − X (t 2 ))] = E[ X (t1 + a ) X (t 2 + a ) + X (t1 ) X (t 2 ) − X (t1 ) X (t 2 + a ) − X (t 2 ) X (t1 + a )] = R X (t1 + a , t 2 + a ) + R X (t1 , t 2 ) − R X (t1 , t 2 + a ) − R X (t1 + a , t 2 )
(3)、令 Z (t ) = aW ( t a 2 ) ⇒ µ Z (t ) = aE[W ( t a 2 )] = 0 C Z (t1 , t 2 ) = E[ aW ( t1 a 2 ) aW ( t2 a 2 )] = a 2 E[W ( t1 a 2 )W ( t2 a 2 )] = a 2σ 2 min{ t1 a 2 , t2 a 2 } = σ 2 min{t1 , t 2 }, t1 , t 2 ≥ 0
2
), a为正常数;
解 (1)、令 Z (t ) = W (t ) + At ⇒ µ (t ) = E[W (t ) + At ] = At Z
C Z (t1 , t 2 ) = E[( Z (t1 ) − µ Z (t1 ))( Z (t 2 ) − µ Z (t 2 ))] = E[(W (t1 ) + At1 − At1 )(W (t 2 ) + At 2 − At 2 )] = E[W (t1 )W (t 2 )] = σ 2 min{t1 , t 2 }, t1 , t 2 ≥ 0
假定 Z (t ) = X + Yt , t ∈ R.若已知二维随机变量 例3 σ 12 ( X , Y )的协方差矩阵为 ρσ 1σ 2 的协方差函数.
ρσ 1σ 2 ,试求 Z (t ) 2 σ2
解 CZ (t1 , t2 ) = E[( X + Yt1 − ( µ X + µY t1 ))( X + Yt2 − ( µ X + µY t2 ))] = E[(( X − µ X ) + (Yt1 − µY t1 ))(( X − µ X ) + (Yt2 − µY t2 ))] = E[( X − µ X )( X − µ X )] + t2 E[( X − µ X )(Y − µY )] +t1 E[(Y − µY )( X − µ X )] + t1t2 E[(Y − µY )(Y − µY )]
= P{ X (t ) ≤ x} = FX ( x, t )
RY (t1 , t2 ) = E[Y (t1 )Y (t2 )] = 1× P {Y (t1 )Y (t2 ) = 1} + 0 × P {Y (t1 )Y (t2 ) = 0} = P {Y (t1 )Y (t2 ) = 1}
= P { X (t1 ) ≤ x1 , X (t2 ) ≤ x2 } = FX ( x1 , x2 ; t1 , t2 )
CY (t1 , t 2 ) = E [(Y (t1 ) − µY (t1 ))(Y (t 2 ) − µY (t 2 ))] = E [( X (t1 ) + ϕ (t1 ) − µ X (t1 ) − ϕ (t1 )) ⋅ ( X (t 2 ) + ϕ (t 2 ) − µ X (t 2 ) − ϕ (t 2 ))] = E [( X (t1 ) − µ X (t1 ))( X (t 2 ) − µ X (t 2 ))] = C X (t1 , t 2 )
随机过程补充例题
随机过程 { X (t ), t ∈ T }, x是任一实数,定义另一随机 例1 1, 过程 Y (t ) = 0 X (t ) ≤ x X (t ) > x , t ∈ T . 试将 Y (t )的均值函数和自
相关函数用 X (t )的一维和二维分布函数表示.
解 µY (t ) = E[Y (t )] = 1× P{Y (t ) = 1} + 0 × P{Y (t ) = 0}
(2)、令 Z (t ) = W (t ) + Xt ⇒ µ Z (t ) = E[W (t ) + Xt ] = 0 C Z (t1 , t 2 ) = E[ Z (t1 ) Z (t 2 )] = E[(W (t1 ) + Xt1 )(W (t 2 ) + Xt 2 )] = E[W (t1 )W (t 2 )] + E[W (t1 ) Xt 2 ] + E[ Xt1W (t 2 )] + E[ Xt1 Xt 2 ] = σ 2 min{t1 , t 2 } + t1t 2 , t1 , t 2 ≥ 0
2 = C XX + t2C XY + t1CYX + t1t2CYY = σ 12 + (t1 + t2 ) ρσ 1σ 2 + t1t2σ 2
例4 设X (t )和Y (t ){t ∈ (0, ∞)}是两个相互独立的、分
别具有强度λ和µ的泊松过程.试证明S (t ) = X (t ) + Y (t ) 是具有强度为λ + µ的泊松过程. 解 (1)、 X (t ), Y (t ){t ∈ (0, ∞ )}是独立增量过程,X (t )和
1 1 1 且 P[ X ( ) = 0] = P[ X ( ) = 1] = 2 2 2
1 1 1 P[ X ( ) = 0] = P[ X ( ) = 1] = ⇒ 2 2 2 x<0 0, 1 1 F ( x; ) = , 0 ≤ x < 1 2 2 x ≥1 1,
−1, 出现 H 1 X (1) = ⇒ P[ X (1) = −1] = P[ X (1) = 2] = 2 2, 出现 T x < −1 0, 1 ⇒ F ( x;1) = , −1 ≤ x < 2 2 x≥2 1,
例7
设随机过程X (t ) = e
− At
, t > 0, 其中A是在区间(0, a )
上服从均匀分布的随机变量,试求X (t )的均值函数和相 关函数.
解
µ X (t ) = E[ X (t )] = E[e
− At
]= ∫e
0
a
− tx
1 ⋅ dx a
1 = (1 − e − at ); (t > 0) at a − At1 − At 2 − x ( t1 + t2 ) 1 R X (t1 , t 2 ) = E[ X (t1 ) X (t 2 )] = E (e e ) = ∫ e ⋅ dx a 0
C X (t1 , t 2 ) = E[( X (t1 ) − E ( X (t1 )))( X (t 2 ) − E ( X (t 2 )))] = E[( X − E ( X )) 2 ] = D ( X ) = σ 2
例9 给定随机过程{ X (t ), t ∈ T }和常数a, 试以X (t )的自
例6 利用抛掷硬币的试验定义一随机过程: cos π t , 出现H ， ∞ < t < +∞， X (t ) = − 出现T 2t ,
1 , 试确定X (t )的(1)一维分布函数F ( x; 1 ), 假设P( H ) = P(T ) = 2 2 1 F ( x;1);(2)二维分布函数F ( x1 , x2 ; ,1); 2 解 0, 出现 H 1 由 X (t )的定义 ⇒ X ( ) = , 2 1, 出现 T
∴{S (t ), t > 0}是强度为λ + µ的泊松过程
设{W (t )，t ≥ 0}是以σ 2为参数的维纳过程，求下列 例5 过程的协方差函数： (1)W (t ) + At , ( A为常数); (2)W (t ) + Xt , X 为与{W (t )，t ≥ 0}相互独立的标准正态变量; (3)aW ( t a
已知随机过程 { X (t ), t ∈ T }的均值函数µ X (t )和协 例2 方差函数C X (t1 , t2 ), ϕ (t )是普通函数.试求随机过程Y ( t ) = X ( t ) + ϕ (t )的均值函数和协方差函数.
解பைடு நூலகம்
µY (t ) = E[ X (t ) + ϕ (t )] = µ X (t ) + ϕ (t )
例10
设X (t ) = At + B, t ∈ R, 其中A、B是相互独立，且
2
均服从N (0，σ )分布的随机变量.试证明X (t )是一正 态过程, 并求出它的相关函数(协方差函数).
解 A、 B是相互独立的正态变量 ⇒ ∀时刻 t1 , t 2 ,L , t n , 有
X (t1 ), X (t 2 ),L , X (t n )均为正态变量 ⇒ ∀n ≥ 1, 有 ( X (t1 ), X (t 2 ),L , X (t n ))为 n维正态随机变量 ⇒ X (t )为正态过程.
k
k
= ∑ P{ X (t ) − X (t0 ) = i}P{Y (t ) − Y (t0 ) = k − i}
i =0 k
[λ (t − t0 )]i − λ ( t − t0 ) [ µ (t − t0 )]k −i − µ ( t − t0 ) =∑ e e i! (k − i )! i =0 e− ( λ + µ )( t− t0 ) k i = Ck [λ (t − t0 )]i ⋅ [ µ (t − t0 )]k −i ∑ k! i =0 [(λ + µ )(t − t0 )]k − ( λ + µ )( t − t0 ) = e ⇒ S (t ) − S (t0 ) π [(λ + µ )(t − t0 )] k!
1 = [1 − e − a ( t1 + t2 ) ]; (t1 , t 2 > 0) a (t1 + t 2 )
例8
设随机过程X (t ) ≡ X (随机变量), E ( X ) = a,
D( X ) = σ 2 (σ > 0), 试求X (t )的均值函数和协方差函数.
解
E ( X ) = a , D ( X ) = σ 2 ⇒ µ X (t ) = E[ X (t )] = E[ X ] = a
Q
X (t ), Y (t )相互独立 ⇒
P{S (t ) − S (t0 ) = k } = P{ X (t ) + Y (t ) − X (t0 ) − Y (t0 ) = k } = ∑ P{ X (t ) − X (t0 ) = i , Y (t ) − Y (t0 ) = k − i}
i=0
Y (t )相互独立 ⇒ {S (t ), t ∈ (0, ∞ )}是独立增量过程
(2)、 S (0) = X (0) + Y (0) = 0
(3)、 ∀ t > t 0 > 0, X (t ) − X (t 0 ) Y (t ) − Y (t 0 )
π ( λ (t − t 0 ))
π ( µ (t − t 0 ))
R X (t1 , t 2 ) = E[ X (t1 ) X (t 2 )] = E[( X (t1 ) − E ( X (t1 )))( X (t 2 ) − E ( X (t 2 ))] = C X (t1 , t 2 ) = E[( At1 + B )( At 2 + B )] = t1t 2 E ( A 2 ) + t1 E ( AB ) + t 2 E ( BA) + E ( B 2 ) = σ 2 (1 + t1t 2 )